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3 years ago

Given the QY’=4.125, what is QY?

Mathematics

1 answer:

Goryan [66]3 years ago

3 0

Answer:

Step-by-step explanation:

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Square ABCD has an area of 1600. E is the midpoint of AC and Fis the midpoint of AB.

ryzh [129]

B……….. thank me later.

4 0

2 years ago

What are the solutions to m2 – 9 = 0?

AysviL [449]

Answer:

The answer is m = 3, -3

Step-by-step explanation:

6 0

3 years ago

7 + (4 + 6) = 7+n name the propeerty

snow_tiger [21]

Answer:

n=15

Step-by-step explanation:

7+(4+6)=7+n

28+6=7+n

22=7+n

22-7=n

15=n

8 0

3 years ago

Drag the similar figure into the table.

Wittaler [7]

There wasn't no pictures nor choices so I'm afraid that I had to look it up... The first link was from Brainly and "Tetianaalnasraween05" was asking the same question.

Original Figure: 6 and 3

'Similar' Figure(s):

1.) 4 and 2

2.) 10 and 4

3.) 6 and 6

Ratio of sides are 3/2 and 6/4, 6/4 could be simplified as 3/2, so as my guess, it is, 4 and 2

____

I hope this helps, as always. I wish you the best of luck and have a nice day, friend..

4 0

3 years ago

Verify csc^4x-cot^4x=2csc^2x-1

Oksanka [162]

First, factor the left side:

$(csc^2(x))^2-(cot^2(x))^2$

This can be simplified using $x^2-y^2 = (x+y)(x-y)$ where the csc term is x and the cot term is y:

$(csc^2(x)+cot^2(x))(csc^2(x)-cot^2(x))$

One of our Pythagorean Identities states that:

$csc^2(x)-cot^2(x)=1$

So the left side is now:

$csc^2(x)+cot^2(x)$

Now we can use the same Pythagorean Identity to change the right side to:

$2csc^2(x)-(csc^2(x)-cot^2(x))$

All we did was express 1 in terms of csc and cot.

Distribute the negative:

$2csc^2(x)-csc^2(x)+cot^2(x)$

Combine like terms:

$csc^2(x)+cot^2(x)$

Now the left and right sides are the same.

8 0

3 years ago